The rate of heat generation in a resistive conductor is equivalent to the electrical power (P) dissipated by it. According to Ohm's law, the relationship between power, potential difference (\(V\)), and resistance (\(R\)) is given by:
\(P = \frac{V^2}{R}\)
This formula shows that the rate of heat generation (\(P\)) is directly proportional to the square of the potential difference (\(V^2\)) and inversely proportional to the resistance (\(R\)).
Let the initial potential difference be \(V_1\) and the initial rate of heat generation be \(P_1\). Using the formula:
\(P_1 = \frac{V_1^2}{R}\)
The question states that the potential difference is doubled. Let the new potential difference be \(V_2\). Therefore:
\(V_2 = 2V_1\)
Let the new rate of heat generation be \(P_2\). Substituting \(V_2\) into the power formula:
\(P_2 = \frac{V_2^2}{R} = \frac{(2V_1)^2}{R}\)
Simplifying the expression for \(P_2\):
\(P_2 = \frac{4V_1^2}{R}\)
Now, compare \(P_2\) with the initial power \(P_1\):
\(P_2 = 4 \times \left( \frac{V_1^2}{R} \right) = 4P_1\)
Thus, when the potential difference is doubled, the rate of generation of heat becomes four times the original rate.
Two electric bulbs marked 25 W - 220 V and 100 W - 220 V are connected in series with 440 V supply. Which of the bulb will fuse?
Two heaters are marked 200V,300W & 200V,600W. If the heaters are connected in series and the combination connected in series and the combination connected to a 200 V dc supply, then which option is true out of the following?